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dmitriy555 [2]
3 years ago
8

What does White Fang’s interaction with the setting reveal?

Physics
1 answer:
Tom [10]3 years ago
6 0
Um what? are their any choices? wait this isnt science is it> 0.0
You might be interested in
What is the phase of the Moon if it . . .
anzhelika [568]

Answer:

Waxing Gibbous

Third quarter

Waning Gibbous

Explanation:

If moon rises at 3:00 pm then the phase of the moon will be "Waxing Gibbous".

This is because, the moon is actually not fully illuminated but has achieved more than half of its full illumination.

If the moon is highest in the sky at sunrise then the phase of the Moon will be the "Third quarter"

This is because of the fact that at this position moon will rise at midnight, thus it will be at the highest point at the time of the sunrise.

If the moon sets at 10:00 am then the phase of the Moon is "Waning Gibbous"

This is because of the fact that at this position the Moon is moving towards becoming new Moon but at the same time,  the moon is illuminated more than its half illumination.

8 0
3 years ago
A 50-kg box is being pushed a distance of 8.0 m across the floor by a force P with arrow whose magnitude is 159 N. The force P w
Georgia [21]

Answer:

Wp = 1,272 J

Wf = -940.8 J

Wn = 0

Wg = 0

Explanation:

  • Applying the definition of work, as the product of the component of the  force applied, along the direction of the displacement, times the displacement, we find that due to the normal force is always perpendicular to the surface, it does no work, as it has not a component in the direction of the displacement, so Wn = 0.
  • As the weight goes directly downward, it has no component in the direction of the displacement either, so Wg = 0.
  • We can get the work done by the force applied P, simply as follows:

        W_{p} = F_{p} * d * cos \theta = 159 N * 8.0 m * cos 0 = 1,272 J (1)

  • Finally, we have the work done by the force of friction that always opposes to the displacement, so it has negative sign.
  • The frictional force , can be written as follows:

        F_{fr} = \mu k * F_{n} (2)

        where μk = coefficient of kinetic friction = 0.24

         Fn = Normal Force

  • In this case, since the surface is level and horizontal, and there is no acceleration of the box in the vertical direction, this means that the normal force (in magnitude) must be equal to the weight:
  • Fn = m*g = 50 kg * 9.8 m/s2 = 490 N
  • Replacing these values in (2), we get:

       F_{fr} = 0.24 * 50 kg* 9.8 m/s2 = 117.6 N

  • Applying the definition of work, we can get the work done by the frictional force, as follows:

        W_{ffr} = F_{fr} * d * cos \theta = F_{fr} * d * cos (180) = - F_{fr} * d = -117.6 N * 8.0 m \\ =  W_{ffr} = -940.8 N

4 0
3 years ago
A car that weighs 1.0 x 10^4 N is initially moving at a speed of 38 km/h when the brakes are applied and the car is brought to a
hram777 [196]

Answer:

Part a) Force on car = 2833.84 Newtons

Part b) Time to stop the car = 3.8 seconds

Part c) Factor for stopping distance is 4.

Part d) Factor for stopping time is 1.

Explanation:

The deceleration produced when the car is brought to rest in 20 meters can be found by third equation of kinematics as

v^2=u^2+2as

where

v = final speed of the car ( = 0 in our case since the car stops)

u = initial speed of the car = 38 km/hr =\frac{38\times 1000}{3600}=10.56m/s

a = deceleration produces

s = distance in which the car stops

Applying the given values we get

0^2=10.56^2+2\times a\times 20\\\\a=\frac{0-10.56^2}{2\times 20}\\\\\therefore a=-2.78m/s^2

Now the force can be obtained using newton's second law as

Force=\frac{Weight}{g}\times a

Applying values we get

Force=\frac{1.0\times 10^4}{9.81}\times -2.78\\\\\therefore F=-2833.84Newtons

The negative direction indicates that the force is opposite to the motion of the object.

Part b)

The time required to stop the car can be found using the first equation of kinematics as

v=u+at with symbols having the same meanings

Applying values we get

0=10.56-2.78\times t\\\\\therefore t=\frac{10.56}{2.78}=3.8seconds

Part c)

From the developed relation of stopping distance we can see that the for same force( Same acceleration) the stopping distance is proportional to the square of the initial speed thus doubling the initial speed increases the stopping distance 4 times.

Part d)

From the relation of stopping time and the initial speed we can see that the stopping distance is proportional initial speed thus if we double the initial speed the stopping time also doubles.

8 0
3 years ago
4. A net force of 15 N is exerted on a book to cause it to accelerate at a rate of 5 m/s2. Determine the mass of the
kenny6666 [7]

Answer:

<h2>3 kg </h2>

Explanation:

The mass of an object given it's force and acceleration can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m =  \frac{15}{5}  = 3 \\

We have the final answer as

<h3>3 kg</h3>

Hope this helps you

4 0
2 years ago
What is water potential???​
melisa1 [442]

Answer:

Water potential is the potential energy of water per unit volume relative to pure water in reference conditions. Water potential quantifies the tendency of water to move from one area to another due to osmosis, gravity, mechanical pressure and matrix effects such as capillary action.

3 0
3 years ago
Read 2 more answers
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